python 嵌套列表排序_在Python中对嵌套列表进行排序和分组

对于第一个问题，你应该做的第一件事是按第二个字段排序列表：

x = [

['4', '21', '1', '14', '2008-10-24 15:42:58'],

['3', '22', '4', '2somename', '2008-10-24 15:22:03'],

['5', '21', '3', '19', '2008-10-24 15:45:45'],

['6', '21', '1', '1somename', '2008-10-24 15:45:49'],

['7', '22', '3', '2somename', '2008-10-24 15:45:51']

]

from operator import itemgetter

x.sort(key=itemgetter(1))

然后你可以使用itertools的groupby函数：

from itertools import groupby

y = groupby(x, itemgetter(1))

现在y是一个包含元素(元素，项迭代器)的迭代器。解释这些元组比显示代码更容易混淆：

for elt, items in groupby(x, itemgetter(1)):

print(elt, items)

for i in items:

print(i)

其中打印：

21

['4', '21', '1', '14', '2008-10-24 15:42:58']

['5', '21', '3', '19', '2008-10-24 15:45:45']

['6', '21', '1', '1somename', '2008-10-24 15:45:49']

22

['3', '22', '4', '2somename', '2008-10-24 15:22:03']

['7', '22', '3', '2somename', '2008-10-24 15:45:51']

对于第二部分，应该使用列表推导，如上所述：

from pprint import pprint as pp

pp([y for y in x if y[3] == '2somename'])

其中打印：

[['3', '22', '4', '2somename', '2008-10-24 15:22:03'],

['7', '22', '3', '2somename', '2008-10-24 15:45:51']]